Discussion Forum

if the function $f:[a,b]\rightarrow \left[-\frac{\sqrt{3}}{4},\frac{1}{2}\right]$ defined by

$f(x)=\begin{bmatrix}1 & 1&1 \\1 & 1+\sin_{}x&1\\1+\cos x&1&1 \end{bmatrix}$

 is one-one and onto , then

Answer:

Explanation:

We have,  $f:[a,b]\rightarrow \left[-\frac{\sqrt{3}}{4},\frac{1}{2}\right]$ 

$f(x)=\begin{bmatrix}1 & 1&1 \\1 & 1+\sin_{}x&1\\1+\cos x&1&1 \end{bmatrix}$

$f(x)=\begin{bmatrix}1 & 1&0 \\0 & \sin_{}x&0\\1+\cos x&-\cos x&-\cos x \end{bmatrix}$

    $[C_{3}\rightarrow C_{3}-C_{1},C_{2}\rightarrow C_{2}-C_{1}]$

 f(x) =-sin x cos x

$f(x)=-\frac{1}{2} \sin 2x$

  $f(x)\in \left[\frac{\sqrt{3}}{4},\frac{1}{2}\right]$

$\therefore$    $-\frac{\sqrt{3}}{4}\leq \frac{-1}{2} \sin 2x \leq \frac{1}{2}$

$\Rightarrow$      $-1\leq  \sin 2x \leq \frac{\sqrt{3}}{2}$

 $\Rightarrow$    $-\frac{\pi}{4}\leq  x \leq \frac{\pi}{6}$

$\therefore$       $a=\frac{-\pi}{4},b=\frac{\pi}{6}$